Set of intersection of elements of a dynkin system with a fixed element is a dynkin system In E. Cinlar Probability and stochastics, there is a short lemma used to prove the Monotone class theorem. 

It is not clear to me as to why with $ A \in \hat{D} ,B \in \hat{D},B \subseteq A \Rightarrow A \setminus B \in \hat{D}$ 
 A: I will answer the question in its entirety. This is Lemma 1.7 (also Exercise 1.14) of Probability and Stochastics, by Erhan Çınlar.
The definition of a d-system is:



Proof of Lemma 1.7.
a) First, $E \in \mathscr D$. Because $D \subseteq E$, we have $E \cap D = D \in \mathscr D$, so $E \in \hat{\mathscr D}$.
b) Now consider $A, B \in \hat{\mathscr D}$ where $B \subseteq A$. We want to show that $A \setminus B \in \hat{\mathscr D}$. That means we want to show that $(A \setminus B) \cap D \in \mathscr D$.
By assumption, we know that $A \cap D \in \mathscr D$, and $B \cap D \in \mathscr D$, and $A \setminus B \in \mathscr D$. So the question seems to be how to combine these facts in order to reach the desired conclusion. I think d-systems are closed under complements but not necessarily under union and intersections, though.
The key seems to be writing $(A \setminus B) \cap D = (A \cap D) \setminus (B \cap D)$, which can easily be verified. This is important because the sets on either side of the slash are elements of $\mathscr D$, and because $B \subseteq A$, we must have $B \cap D \subseteq A \cap D$. Therefore, by the second property of a d-system we conclude that $(A \cap D) \setminus (B \cap D) = (A \setminus B) \cap D \in \mathscr D$, and therefore $A \setminus B \in \hat{\mathscr D}$ as desired.
c) So we have an increasing sequence $(A_n)$ in $\hat{\mathscr D}$, and we want to show that $\cup_{i = 1}^{\infty} A_i \in \hat{\mathscr D}$.
We have $A_1 \cap D \in \mathscr D, A_2 \cap D \in \mathscr D, A_3 \cap D \in \mathscr D, \dots$ and furthermore, $A_1 \cap D \subseteq A_2 \cap D \subseteq A_3 \cap D \subseteq \dots$\ . We therefore have a second increasing sequence, call it $(A_n \cap D)$, in $\mathscr D$, so the limit $\cup_{i = 1}^{\infty} (A_i \cap D)$ is in $\mathscr D$. But we can rewrite $\cup_{i = 1}^{\infty} (A_i \cap D) = (\cup_{i = 1}^{\infty} A_i) \cap D \in \mathscr D$, which tells us that $\cup_{i = 1}^{\infty} A_i \in \hat{\mathscr D}$ as desired. Q.E.D.
(Note, the script D is displayed differently on this site than it is in the text.)
